Question: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $n \neq 0$. $r = \dfrac{6(4n + 9)}{9} \div \dfrac{40n + 90}{8n} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{6(4n + 9)}{9} \times \dfrac{8n}{40n + 90} $ When multiplying fractions, we multiply the numerators and the denominators. $r = \dfrac{ 6(4n + 9) \times 8n } { 9 \times (40n + 90) } $ $ r = \dfrac {8n \times 6(4n + 9)} {9 \times 10(4n + 9)} $ $ r = \dfrac{48n(4n + 9)}{90(4n + 9)} $ We can cancel the $4n + 9$ so long as $4n + 9 \neq 0$ Therefore $n \neq -\dfrac{9}{4}$ $r = \dfrac{48n \cancel{(4n + 9})}{90 \cancel{(4n + 9)}} = \dfrac{48n}{90} = \dfrac{8n}{15} $